Assignment # 2

Fall 2002 Semester

Second Degree Equation

**Problem
2- 6 : Graph**

**y
= 2x ^{2 }+ 3x - 4**

**i.
Overlay
a new graph replacing each x with (x 4).**

**ii.
Change
the equation to move the graph into the second quadrant.**

**iii.
Change
the equation to produce a graph concave down that shares the same vertex.**

**The
above graph is the representation for the initial equation: y = 2x ^{2} +
3x -4.**

**There
is one y-intercept point (when x = 0) of -4.
There are 2 solutions for x (when y =0) of ~0.85 and ~-2.35.**

**The
vertex of the parabola created, is determined to be at x = -0.75 and y = -5.125.**

**i.
Overlay
a new graph replacing each x with (x 4). **

**Graph
of overlay of initial equation with equation replacing x by x -4.
The equations plotted are:**

**2x ^{2} + 3x -4 (Green)**

**and**

**2(x-4) ^{2}
+ 3(x-4) -4 (Red)**

**
**

**The
net affect of changing from x to x-4 is to shift the parabola
further up the x-axis. All of
the x parameter items from the initial graph are increased by 4. e.g.
the vertex of the parabola is now at x = 3.25, y = 5.125.**

**ii
. Moving the graph into the 2 ^{nd}
quadrant**

**The
problem here is to adjust the parameters to provide a graph that appears in
quadrant where:**

**y
> 0**

**and**

**x
< 0**

**From
step i. it appears that all that 2 things need to be done:**

**Shift the parabola to the left by using x+n where the value of n will allow the parabola to be far enough left to allow for the plot not to cross the y-axis.****Change the vertex point for y by increasing the intercept parameter e.g. -4 to +1\2**

**e.g.
y = 2(x+100) ^{2} + 3(x+100) + 2**

**However,
it appears that this will not work, as any increase in the value of n will
eventually cross the y-axis (although usually at some point well beyond where
were able to plot. For the
purposes of demonstration, it appears to be within quadrant 2.**

**
**

**The
graph below changed the equation to the following:**

**y = e ^{|(-x)
-10|} + 3(|-x-10|) + 1**

**This
also appears to be within the 2 ^{nd} quadrant, however, as with the
first, when y is large enough, there is an intersection with the y-axis (there
is a solution for y when x is zero or positive.)**

**
**

**iii.
The
following graph is the result of changing the equation to produce a graph
concave down that shares the same vertex with the first equation.
The equations used are as follows:**

** 2x ^{2}
+ 3x -4 (Green)**

**-2x ^{2}
- 3x -6.25 (Red)**

The second equation was determined by:

· Determining the vertex point of the original graph.

· Inverting the original graphs parabola.

· Changing the y intercept parameter (-4) to match the inverted parabola.

__Vertex
Point Determination:__

- From calculus, note that the minimum and maximum points may be determined from the first derivative of the equation. The first derivative provides us with a linear equation that represents the slope of the curve at any of the points. The minimums and maximums of the equation would be points where the slope would be zero. i.e.

First
derivative of: y = 2x^{2} +
3x 4 yields ΰ
y = 4x + 3.

- Next determine the value of x when y = 0:

When y = 0, the x = -0.75

- Then calculate y for when x = -0.75

For
y = 2x^{2} + 3x 4, when x = -0.75, then y = -5.125

__Inverting
the Parabola:__

The
parabola can be inverted, by changing the polarity of the x-parameter parts of
the equation. i.e. y = 2x^{2}
+ 3x 4 becomes -2x^{2} - 3x 4.

This yields the graph below. Still not finished.

__Changing
the y intercept parameter (-4) to match the inverted parabola:__

The y intercept needs to be changed for the new parabola. We determined the vertex point for the original graph to be at x = -0.75 and y = -5.125. These same vertex points apply to the new calculation. What is needed is to replace -4 by a variable name (e.g. n), and substitute the x and y values for the vertex (-0.75 and -5.125, respectively).

i.e.
-5.125 = -2(-0.75)^{2} -3(-0.75) n;

n = 5.125 1.125 + 2.25 ΰ 6.25

**The
final equation becomes: -2x ^{2} - 3x - 6.25 **